Understanding the Schwarz reflection principle

$f(\bar{z})$ is not holomorphic (unless $f$ is constant), as $d(f(\bar{z}))=f'(\bar{z}) d\bar{z}$ is not a multiple of $dz$ (but rather of $d\bar{z}$). Perhaps more intuitively: holomorphic = conformal and orientation-preserving; $f(\bar{z})$ is conformal, but changes the orientation (due to the reflection $z\mapsto\bar{z}$). Hence your function $F$ is not holomorphic on $\Omega^-$.

On the other hand, $\overline{f(\bar{z})}$ is holomorphic, as there are two reflections. If $f$ is real on $I$ then by gluing $f(z)$ on $\Omega^+$ with $\overline{f(\bar{z})}$ on $\Omega^-$ you get a function continuous on $\Omega^+\cup I\cup\Omega^-$ and holomorphic on $\Omega^+\cup\Omega^-$. It is then holomorphic on $\Omega^+\cup I\cup\Omega^-$ e.g. by Morera theorem.

You left out the condition that $f$ takes real values on $I.$ Pretend that $0 \in I,$ it changes nothing. Once we succeed in extending to a holomorphic $G$ on both sides of the real axis, this says that the power series of $G$ around $0$ has all real coefficients, if for no better reason than that all derivatives of $G$ at $0$ are real. In turn, this says that $$ G( \bar{z}) = \overline{G(z)} $$